Thanks bltizo. Based in this photo comparition,i have to say that i doubt that type 055 will have 12000t. This ship has aprox. 172+- meters of lenght (compared to 156 meters type 052d) and aprox 21+ meters beam (18 meters type 052d). Type 052d has 7500t. Will type 055 get to 12000t ? i have doubts.
Do you know what the draft value for 055 will be? Otherwise, I don't know how you can rule out a 12k displacement.
Considering Burke FIIA's dimensions (155m x 20m x 9.3m) and displacement (9250 metric tons), with other factors being roughly comparable, it's perfectly within reason that 055 would displace close to 12,000 metric tons, assuming a length of 175m, beam of 21m, and draft of 9.8m (only 5% greater than that of Burke IIA).
Here's my back-of-the-envelope calculation:
[(175mx21mx9.8m)/(155mx20mx9.3m)]x9,250tons=11,555tons
Even a minute increase of the beam and/or draft value would send 055's displacement well north of 12,000 tons.
OR, if you prefer, let's compare 055 to 052D, and I'm gonna make the ludicrously conservative assumption that 055's draft value is EQUAL to Burke IIA's. The calculation is as follows:
[(175mx21mx9.3m)/(156mx18mx6.5m)]x7500tons=14,044tons
A whopping 14,044 tons!
Now, if we change the draft value to a more reasonable 9.8m, we get 14,799 tons. Just for kicks, plug in 21.5m for beam and 10m for draft, we get 15,460 tons.
That's 15,460 tons!
In fact, I'd be surprised if 055 doesn't end up displacing more than 12,000 tons if the rumored dimensions hold up.
Of course, my calculations are based on the assumption that 055 will exist in a three-dimensional space. If you believe the said assumption is unreasonable, please explain why this is so.